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\begin{document}
\parindent=0cm
\section*{ Effect of Mappings }

$ \varphi : M \rightarrow N, \;\; \mathcal{C}^\infty $
$\;\Rightarrow \;\; \varphi_* : TM \rightarrow TN \;\; bundle\;
map.$\\
$And\; \;\varphi_{*}|_{p} \;:\; T_pM \rightarrow T_{\varphi(p)}N,
\;\; linear$ $\;\Rightarrow (\varphi_*|_p)^*\; :
\;T^*_{\varphi(p)}N \rightarrow
T^*_{p}M$, dual map. \\
$This \; induces \; a\; linear\; map \;\;\varphi^*_p\;:=\;
\bigwedge^r(\varphi_*|_p^*)\;: \;\bigwedge^r(T^*_{\varphi(p)}N)
\rightarrow \bigwedge^r(T^*_{p}M)$\\

$For \; \alpha \in \mathcal{E}^1(N),\; \varphi^*\alpha \in
\mathcal{E}^1(M) \;is \; defined \;\;as$ $\varphi^*\alpha|_p =
\varphi^*_p(\alpha|_{\varphi(p)}).$
\\
Note that $\varphi^*\alpha(X_p)\; = \; \alpha(\varphi_*X_p) $\\


Similarly, for $\alpha \in \mathcal{E}^r(N),\; \varphi^*\alpha \in
\mathcal{E}^r(M)$ \; and
\begin{eqnarray*}
\varphi^*\alpha(X_1, \cdots, X_r)(p)\; &=&
\;\varphi^*\alpha(X_1(p), \cdots, X_r(p))\\
&=&
\alpha((\varphi_*X_1)(\varphi(p)),\cdots,(\varphi_*X_r)(\varphi(p)))\\
&=& \alpha(\varphi_*X_1,\cdots,\varphi_*X_r )(\varphi(p))
\end{eqnarray*}

$\Rightarrow\; \varphi^*\; : \; \mathcal{E}(N) \; \rightarrow \;
\mathcal{E}(M)=\displaystyle{\bigoplus_{r=0}^{\infty}}\mathcal{E}^r(M).$\\

이 경우 differential $d$에 관하여 $\mathcal{E}(M)$은 "differential
graded algebra"가 된다.

\begin{prop}
$Let\;\; \val \; : \; M \; \rightarrow \; N $ , \; $\ccn.$ Then
$\val^*\; : \; \mathcal{E}(N)\; \rightarrow \; \mathcal{E}(N)$ is
a differential
graded algebra homomorphism, i.e., \\\\
$1.\; \val^*(\alp+\bet)\; = \; \val^*\alp \; + \; \val^*\bet $\\
$2.\; \val^*(\alp\wedge\bet)\; = \; \val^*\alp \;\wedge \;
\val^*\bet
$\\
$3.\; \val^*d\; = \; d\val^* $
\end{prop}

\begin{proof}
1 and 2 are clear.\\
$(\because)\; f\;:\; V\; \rightarrow \; W \; \Rightarrow \;
f^*\;:\;W^*\;\rightarrow\;V^* $\\
$\Rightarrow\; \bigwedge(f^*)\; : \; \bigwedge(W^*)\;
\rightarrow \; \bigwedge(V^*) $ is an algebra homomorphism.\\
And note that $+, \wedge$\; are pointwise operations.\\

Show $\val^*\alp$ is $\ccn$ if $\alp\; \in \mathcal{E}^r(N):$\\
1st $\val^*(df)$ case :\\
$\val^*(df)(X)(p)\; = \; \val^*(df)(X_p)\; = \; df(\val_*X_p)$


$\hspace{6.33em}=(\val_*X_p)f = X_p(f\circ\val)=X(f\circ\val)(p)
\;\; : \ccn\;-\; function \;\;of\;\; p.$

Moreover, $\hspace{1.7em}=\;d(f\circ\val)(X)(p)$\\
$\therefore \; \val^*(df) \; = \; d(f\circ\val)\; = \; d(\val^*f)$
and$\;\val^*d \; = \; d\val^* \; on \; \mathcal{F}$\\\\\\
$2nd \;\; \alp \; \in \; \mathcal{E}^r(N)$ case:\\
Locally $\val\;:V \rightarrow \; (U,x)$

$\hspace{5em}p \;\; \mapsto \;\val(p)$ \\
Let $\alp\;=\;\sum f_Idx_I$. Then
$\val^*\alp\;=\;\sum\val^*(f_I)\val^*(dx_{i_1})\wedge\cdots \wedge
\val^*(dx_{i_r})\;\; : \ccn.$\\\\
Now we can prove 3 as follows.


$d(\val^*\alp) \; = \; d(\sum (\val^*f_I) d(\val^*x_{i_1})\wedge
\cdots \wedge d(\val^*x_{i_r})) $\\
$=\sum d(\val^*f_I)\wedge d(\val^*x_{i_1})\wedge \cdots \wedge
d(\val^*x_{i_r}) $\\
$=\val^*(\sum df_I \wedge dx_{i_1} \wedge \cdots \wedge
dx_{i_r})\;=\; \val^*d\alp$
\end{proof}

\end{document}
